In the trapezoidal ABCD, point P starts from point a and moves along the edge of ad to point d at a speed of 1cm / s, and point Q starts from point C and moves along CB to point B at a speed of 2cm / s. if points P and Q start from two points at the same time, when one of them reaches the end point, the other stops moving. (1) what is the value of T when trapezoidal pbqd is a parallelogram? (2) What is the value of T when trapezoid pbqd is isosceles trapezoid?

(1) When PD = BQ, trapezoidal pbqd is a parallelogram. According to the meaning of the title, 18-t = 21-2t, the solution is t = 3, that is, when t = 3, trapezoidal pbqd is a parallelogram. (2) PE ⊥ BC and DF ⊥ BC are e and f respectively

In trapezoidal ABCD, if the parallel BC angle B = 50 degrees, the parallel BC angle c = 80 degrees, ad = 10cm, BC = 18cm, then CD =?

CD = 8 cm, extend Ba and CD to point E, because AD / / BC, so angle ead = angle B = 50 degrees, angle EDA = angle c = 80 degrees, then angle BEC = 50 degrees, so triangle ead and triangle EBC are isosceles triangle, that is ed = ad = 10 cm, EC = BC = 18 cm, so CD = ec-ed = 8 cm

If AB = 8, BC = 6, CD = 2, the bisector of ∠ B intersects EF at g, then the length of FG is______ ．

∵ EF is the median line of ladder shaped ABCD, ∵ EF ∥ Ba, EF = 12 (CD + AB) = 5, be = 12bc = 3, ∵ EGb = ∵ GBA, ∵ BG bisection ∵ CBA, ∵ CBG = ∵ GBA, ∵ EGb = ∵ EBG, ∵ eg = be = 3, ∵ FG = 5-3 = 2, so the answer is: 2

As shown in the figure, in ladder ABCD, ad ‖ BC and EF pass through the intersection o of the diagonal of the ladder, and ef ‖ ad. (1) verify: OE = of, (2) calculate the value of oead + oebc; (3) verify: 1AD + 1BC = 2ef

(1) ∵ EF ∥ ad, ad ∥ BC, ∥ oebc = AOAC = ODBD = ofBC, so OE = of; (2) ∵ EF ∥ ad, ad ∥ BC, ∥ oead = BEAB, oebc = aEAb, ∥ oead + oebc = AE + BEAB = ABAB = 1; (3) from (2), OE (1AD + 1BC) = 1, OE = of = 12ef, ∥ 2oeef = 1, ∥ OE (1AD + 1BC) = 2oeef, ∥ 1AD + 1BC = 2ef

As shown in the figure, in isosceles trapezoid ABCD, AB is parallel to CD, AB is greater than CD, ad is equal to BC, diagonal AC and BD intersect at point O, ∠ AOB is equal to 60 ° and EF is respectively
And EF is the midpoint of the point odoa and M is the midpoint of BC respectively
On the transition of summer vacation

Take the midpoint P of OC, the midpoint Q of ob, and connect EP and FQ respectively. From the topic meaning, △ AOB and △ COD are equilateral triangles, then OC = od = CD, OA = ob = ab. e, F, P and Q are respectively the midpoint of OD, OA, OC and OB, which are PE = 1 / 2CD = 1 / 2od = OE, PM = 1 / 2ob = 1 / 2oa = of

In trapezoidal ABCD, ad is parallel to BC, diagonal AC and BD intersect at O, EF is parallel to ad, AB intersects at O, CD intersects at EF
What's the relationship between OE and of? Why 2 find out the value of OE over od plus of over BC

1、
The relationship between OE and of is equal
According to ad ‖ BC, AO / OC = Bo / OD,
Then: AO / (AO + OC) = Bo / (Bo + OD), namely: AO / AC = Bo / BD;
From EF ∥ ad, we can get: EF ∥ BC,
Then: OE / BC = AO / AC = Bo / BD = of / BC,
So OE = of
2. (the ratio of OE to od in the question should be changed to that of OE to ad, otherwise the result is not fixed.)
From EF ‖ ad, OE / ad = do / BD,
So OE / AD + of / BC = do / BD + Bo / BD = (do + Bo) / BD = BD / BD = 1

As shown in the figure, in trapezoidal ABCD, ad ∥ BC, BD and AC intersect at O, and the straight line passing through o intersects AB, CD at e and f respectively, and ef ∥ BC, if ad = 12, BC = 20, then EF=______ ．

∵ EF ∥ ad ∥ BC, ∥ OAD ∥ OCB, OA: OC = ad: BC = 12:20 ∥ OAE ∥ caboe: BC = OA: CA = 12:32 ∥ EF = 2 × 1232 × 20 = 15, so the answer is: 15

In rectangular ABCD, de intersects AB with e BF intersects DC with F. given De is parallel to BF, AB is twice as much as AD. calculate AE ratio ab

You are short of conditions in this question. I suggest you check it or take a screenshot of the original question. Good luck!

As shown in the figure, in rectangular ABCD, f is a point on the edge of BC, the extension line of AF intersects the extension line of DC at g, de ⊥ Ag at e, and de = DC

It is proved that: ∵ quadrilateral ABCD is a rectangle, ∵ AB = CD, and ∵ de = DC, ∵ AB = De, ∵ ad ∥ BC, ∵ BFA = ∥ DAE, ∥ in △ ABF and △ DEA, ∥ BFA = ∥ DAE ∥ B = ∥ DEA = 90 ° AB = De, ≌ Abf ≌ DEA, ≌ AE = BF

Point EF is on the sides BC and ab of rectangle ABCD, BF = 3, be = 4, CE = 3, AE and CF intersect at point P, and angle APC = angle AEB + angle CFB, then calculate the area of ABCD

Through e, eg ‖ CF intersects BF with G. ∵ ABCD is a rectangle, ∵ B = 90 ° and BF = 3, BC = be + CE = 4 + 3 = 7, ∵ Tan ∵ BCF = BF / BC = 3 / 7. ∵ APC ∵ EPF, ∵ EPF ∵ AEB ∵ CFB ∵ B = 360 degree, ∵ APC ∵ AEB ∵ CFB ∵ 90 ∵ 360 degree and ∵ APC ∵ AECB ∵ CFB ∵ 90 ∵ 360 degree